After characterising connected subsets of the real line, we now introduce a compact test object for disconnectedness. The discrete two-point space records whether a continuous map can assign two separated labels to a metric space. Two-point tests for connectedness are powerful because they convert a topological decomposition into a continuous function. A common mistake is to forget the word onto in the disconnectedness test: a constant map to the two-point space does not split the domain. The focus keyword two-point tests for connectedness will be used to express connectedness through continuous maps. :::definition[Discrete Two-Point Space] Let $X_0=\{0,1\}$. Define $d_0$ on $X_0$ by $$\begin{equation*} d_0(x,y)= \begin{cases} 0, & x=y,\\ 1, & x\neq y. \end{cases} \end{equation*}$$ Then $(X_0,d_0)$ is called the $discrete two-point space$. Every subset of $X_0$ is open. ::: :::theorem[Disconnectedness Through the Discrete Two-Point Space] Let $(X,d)$ be a metric space. Then $X$ is disconnected if and only if there exists a continuous mapping from $X$ onto the discrete two-point space $X_0$. ::: :::proof Given that $(X,d)$ is a metric space. To prove that $X$ is disconnected if and only if there exists a continuous onto mapping from $X$ to $X_0$. First assume that $X$ is disconnected. Then there exist two nonempty disjoint open subsets $A$ and $B$ of $X$ such that $X=A\cup B$. Define $f:X\to X_0$ by $$\begin{equation*} f(x)= \begin{cases} 0, & x\in A,\\ 1, & x\in B. \end{cases} \end{equation*}$$ Since $A$ and $B$ are nonempty, $f$ is onto. Also, $$\begin{equation*} f^{-1}(\{0\})=A,\qquad f^{-1}(\{1\})=B. \end{equation*}$$ The remaining open subsets of $X_0$ are $\varnothing$ and $X_0$, whose inverse images are $\varnothing$ and $X$. Therefore the inverse image of every open subset of $X_0$ is open in $X$. Hence $f$ is continuous. Conversely, assume that there exists a continuous onto mapping $f:X\to X_0$. Let $$\begin{equation*} A=f^{-1}(\{0\}),\qquad B=f^{-1}(\{1\}). \end{equation*}$$ Since $f$ is onto, both $A$ and $B$ are nonempty. Since $f$ is continuous and $\{0\}$ and $\{1\}$ are open in $X_0$, both $A$ and $B$ are open in $X$. Also $X=A\cup B$ and $A\cap B=\varnothing$. Therefore $X$ is disconnected. $\square$ ::: :::corollary Let $(X,d)$ be a metric space. Then $X$ is connected if and only if every continuous mapping $f:X\to X_0$ is constant. ::: :::proof Given that $(X,d)$ is a metric space. To prove that $X$ is connected if and only if every continuous mapping from $X$ to $X_0$ is constant. First assume that $X$ is connected. Let $f:X\to X_0$ be continuous. If possible let $f$ be nonconstant. Then both $0$ and $1$ are values of $f$, so $f$ is onto $X_0$. By the preceding theorem, $X$ is disconnected. A contradiction. Therefore $f$ is constant. Conversely, assume that every continuous mapping from $X$ to $X_0$ is constant. Suppose, if possible, that $X$ is disconnected. Then there exists a continuous onto mapping from $X$ to $X_0$. Such a mapping cannot be constant because it is onto a two-point set. A contradiction. Hence $X$ is connected. $\square$ ::: :::example[Indicator of a Separation] Let $X=(-\infty,0)\cup(0,\infty)$ with the usual subspace metric. Define $f:X\to X_0$ by $f(x)=0$ for $x<0$ and $f(x)=1$ for $x>0$. The two inverse images are open in $X$, so $f$ is continuous and onto. Hence $X$ is disconnected. $\square$ ::: :::remark The discrete two-point space detects disconnectedness. A metric space is connected exactly when no continuous function can separate it into two distinct discrete values. ::: :::solved-problem Let $X$ be connected and let $f:X\to\{0,1\}$ be continuous, where $\{0,1\}$ has the discrete metric. Show that $f^{-1}(\{0\})$ is either $\varnothing$ or $X$. ::: :::solution Given that $X$ is connected and $f:X\to X_0$ is continuous. To prove that $f^{-1}(\{0\})$ is either $\varnothing$ or $X$. By the two-point connectedness criterion, every continuous map from $X$ to $X_0$ is constant. If $f$ is constantly $0$, then $f^{-1}(\{0\})=X$. If $f$ is constantly $1$, then $f^{-1}(\{0\})=\varnothing$. Therefore $f^{-1}(\{0\})$ is either $\varnothing$ or $X$. $\square$ ::: :::exercise [1] Define the discrete two-point space. [2] Explain why every subset of $X_0$ is open. [3] State the two-point test for disconnectedness. [4] Show that a nonconstant continuous map from a connected space to $X_0$ cannot exist. ::: :::answer [1] It is $X_0=\{0,1\}$ with $d_0(x,y)=0$ if $x=y$ and $d_0(x,y)=1$ if $x\neq y$. [2] Both singletons are open balls in the discrete metric. [3] $X$ is disconnected if and only if there is a continuous onto map $X\to X_0$. [4] A nonconstant map to $X_0$ is onto, which would disconnect the domain. ::: :::faq[Frequently Asked Questions] Q: Why must the map be onto? A: Onto means both labels occur, so the domain is actually split into two nonempty pieces. Q: Is every function into $X_0$ continuous? A: Not from every space. Continuity requires the inverse images of $\{0\}$ and $\{1\}$ to be open. Q: What is the main use of this theorem? A: It converts separation problems into problems about continuous maps. ::: :::call-to-action[Continue to Continuous Images] subtitle: Apply two-point tests for connectedness to prove that continuous images of connected spaces remain connected. button: Next Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/continuous-images-and-intermediate-values button-ghost: Previous Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/intervals-and-connected-subsets-of-the-real-line :::